Given that the parabola y = ax2-2x + 1 has no intersection with the X axis, then the quadrant of the vertex of the parabola is () A. Fourth quadrant B. third quadrant C. second quadrant D. first quadrant

Given that the parabola y = ax2-2x + 1 has no intersection with the X axis, then the quadrant of the vertex of the parabola is () A. Fourth quadrant B. third quadrant C. second quadrant D. first quadrant

∵ the parabola y = ax2-2x + 1 has no intersection with the x-axis, ∵ △ = 4-4a < 0, the solution is: a > 1, ∵ the opening of the parabola is upward, and ∵ B = - 2, ∵ B2A > 0, ∵ the symmetry axis of the parabola is on the right side of the y-axis, and the vertex of the parabola is in the first quadrant

Given that the parabola y = a (x-1) 2 + H (a ≠ 0) intersects the X axis at two points a (x1, 0) and B (3, 0), then the length of line AB is ()
A. 1B. 2C. 3D. 4

The vertex coordinates (1, H) of ∵ quadratic function y = a (x-1) 2 + H, - B2A = 1, then - BA = 2 and ∵ x2 = 3 ∵ X1 + x2 = X1 + 3 = 2, the solution is X1 = - 1 ∵ the length of AB = | x1-x2 | = | - 1) - 3 | = 4

It is known that the parabola y = x2 + (1-2a) x + A2 (a ≠ 0) intersects the X axis at two points a (x1,0) and B (x2,0) (x1 ≠ x2)
(1) Find the value range of a and prove that a and B are on the left side of origin o;
(1) ∵ the parabola and X-axis intersect at two points a (x1,0), B (x2,0), and x1 ≠ x2,
∴△=（1-2a）2-4a2＞0．a＜ 14．
And ∵ a ≠ 0,
∴x1•x2=a2＞0,
That is, X1 and X2 must have the same number
And X1 + x2 = - (1-2a) = 2a-1 < 2 / 4-1 = - 1 / 2 < 0,
Both X1 and X2 must be negative,
The points a (x1,0) and B (x2,0) are all on the left side of the origin
I want to know how to find X1 + x2 = - (1-2a) = 2a-1 < 2 / 4-1 = - 1 / 2 < 0,
How did you get it?
Especially 2 / 4-1 = - 1 / 2

If the parabola y = ax ^ 2 + BX + C intersects with X axis x1, X2, then:
X1 + x2 = - B / A, which is the relationship between the root and the coefficient, that is, the Veda theorem, similar to the quadratic equation;
The above result is: 1-4a > 0, so a

What is the smallest even number divisible by 2 / 3 / 5

2 * 3 * 5 = 30
Since 2, 3 and 5 are prime numbers, 30 is the least common multiple, so
That's 30

Negative 300 times negative 1 / 8 plus 0.125 times 54.5 plus negative 1 / 5 times negative 12.5!

0.15

First simplify: 4-A 2A2 + 6A + 9 △ a-22a + 6 + 2, and then choose any number you like to evaluate

The original formula = (2 + a) (2-A) (a + 3) 2 · 2 (a + 3) A-2 + 2 (1 point) = - 2A + 4A + 3 + 2A + 6A + 3 (2 points) = 2A + 3 (3 points) a any number other than - 3 and 2 can be given if the calculation is correct. (5 points)

How many groups of 1-33 numbers can you choose 6 numbers to add up to 63
Do you have any specific figures

21 groups

If x + 4 = y2x − y = a, X and y are opposite to each other, then a=______ ．

From the meaning of the question: x + y = 0, that is y = - x, substituting into the equations: x + 4 = − X2X + x = a, the solution: a = - 6, x = - 2

X + y = 28 3Y + Z = 810 2A + 2Y = 214 x + a = 19 find the value of X, y.z.a

X + y = 28 (1) 3Y + Z = 810 (2) 2A + 2Y = 214 (3) x + a = 19 (4) x + y = 28 (4) multiply by 2 to get: 2x + 2A = 38 (5) (5) - (3) get: 2x-2y = - 176 (6) (1) multiply by 2 to get: 2x + 2Y = 56 (7) (6) + (7) get: 4x = - 120 x = - 30, y = 58, z = 636, a = 49

It is known that a = LG (1 + 1 / 7), B = (1 + 1 / 49). The algebraic expression of a and B is used to express lg1.4
1/7(a-4b+1)

B = LG (1 + 1 / 49): a = LG (1 + 1 / 7) = LG8 / 7 = lg8-lg7 = 3lg2-lg7, B = LG (1 + 1 / 49) = LG50 / 49 = lg50-lg49 = LG100 / 2 - lg49 = lg100-lg2-lg7 ^ 2 = 2-lg2-2lg7, because 3lg2-lg7 = a2-lg2-2lg7 = B, the two formulas are combined to get LG2 = (2a-b + 2) / 7lg7 = (6-a-3b) / 7lg1.4 = lg14 / 10 =